What is the ratio of the wavelengths of radia | vedclass

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(B) The wavelength $\lambda$ of radiation emitted during an electronic transition is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac

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$20 : 27$

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(B) The wavelength $\lambda$ of radiation emitted during an electronic transition is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$,where $R$ is the Rydberg constant.For the transition from $n_2 = 4$ to $n_1 = 2$: $\frac{1}{\lambda_1} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} ight] = R \left[ \frac{1}{4} - \frac{1}{16} ight] = R \left[ \frac{4-1}{16} ight] = \frac{3R}{16}$. Thus,$\lambda_1 = \frac{16}{3R}$.For the transition from $n_2 = 3$ to $n_1 = 2$: $\frac{1}{\lambda_2} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left[ \frac{9-4}{36} ight] = \frac{5R}{36}$. Thus,$\lambda_2 = \frac{36}{5R}$.The ratio of the wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{16}{3R} 	imes \frac{5R}{36} = \frac{16 	imes 5}{3 	imes 36} = \frac{80}{108} = \frac{20}{27}$.Therefore,the ratio is $20 : 27$.

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

What is the ratio of the wavelengths of radiations emitted when an electron in a hydrogen atom jumps from the fourth orbit to the second orbit and from the third orbit to the second orbit?

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